(1) If C and D are categories and there is a forgetful functor U:C→D, then a C-ification functor F:D→C is an adjoint to U. For example, the (left) adjoint to the forgetful functor from groups to monoids is "groupification" of a monoid, given by formally adjoining inverses. The (left) adjoint to the forgetful functor from presheaves to sheaves is the usual "sheafification" functor.

Note that whenever you have a (left adjoint) C-ification functor F (whenever you have an adjunction, for that matter), you get a universal property. For any object X∈D, there is a canonical morphism (called the unit of adjunction) εX:X→U(F(X)) with the property that any morphism f:X→U(Y) factors as f=U(g)\circ εX for a unique morphism g:F(X)→Y in C.

(2) A scheme X is separated if the diagonal morphism X→XxX is a closed immersion. It is enough to check that the image of the diagonal is closed. Being separated is the algebro-geometric analogue of being hausdorff, which nothing in algebraic geometry ever is.

My question is whether there exists a "separification" functor adjoint to the forgetful functor U from the category of separated schemes to the category of schemes. Note that the forgetful functor U does not respect colimits (you can glue together separated schemes to get a non-separated scheme), so it has no hope of having a right adjoint. But U does respect limits (it's enough to show that an arbitrary product of separated schemes is separated and that fiber products of separated schemes are separated), so it might have a left adjoint.

To put it another way, given a scheme X, is there a canonically defined separated scheme Xs and a morphism X→Xs so that any morphism from X to a separated scheme factors uniquely through X→Xs?

Related questions I'd like to know the answer to:

Is there a "relative separification" functor. That is, does an arbitrary morphism of schemes f:X→Y admit a canonical factorization through a separated morphism fs:X'→Y. This would be analogous to Stein factorization, which I regard as "relative affinification". An arbitrary (quasi-compact and quasi-separated) morphism f:X→Y canonically factors through the affine morphism SpecY(f*OX)→Y

Is there a separification functor for algebraic spaces? Is it possible that the separification of a scheme is naturally an algebraic space?

Is there a separification functor for algebraic stacks? (An algebraic stack is separated if the diagonal is proper.)

I like "separification". I've also heard people propose "sheafication" instead of "sheafification".
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Anton GeraschenkoOct 18 '09 at 2:20

Your proof that $U$ respects limits does not work. Here is a different approach: Since affine schemes are separated, it is enough to prove the following If $\{X \to X_i\}$ is a cone of schemes such that for all affine schemes $T$ the induced map $Hom(T,X) \to lim_i Hom(T,X_i)$ is bijective, then this is already true for all schemes $T$. But this is easy because both sides are sheaves in $T$ with respect to the Zariski Topology.
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Martin BrandenburgJun 14 '11 at 15:29

4 Answers
4

I think that it is highly unlikely that there exists a separification functor. What does exist is the following:

Theorem (Raynaud-Gruson): Let S be a base scheme and work relative to S. Given a non-separated scheme X of finite type, there is a blow-up (a proper birational morphism) X'->X such that X' admits an étale morphism to a projective scheme Z (in particular a separated scheme).

Note that there are non-separated schemes which does not even admit a quasi-finite morphism onto a separated scheme (e.g. take A^2 with a double origin and blow-up one of the origins).

The Theorem is false as stated for non-locally separated algebraic spaces. There are 3 different solutions to this:

Given a scheme $X$, consider the category of separated schemes $Y$ together with a morphism $X \to Y$. View this as a diagram, and take the inverse limit. This means we take the product of all schemes $Y$, and set $X^s$ equal to the intersection of the closed graphs of each arrow in this category.

Then every morphism from $X$ to a separated scheme factors through $X^s$. However this is unique because, were there two different morphisms, there equalizer would be a nonempty closed subscheme of $X^s$ which itself is separated and admits a map from $Y$, which is impossible because $X^s$ would have to lie in a map of this graph.

Then we get functoriality the obvious way.

The only issue is set-theoretic. However somehow I think we only need consider maps to separated schemes of bounded cardinalities.

There's likely some obvious flaw in this thinking. If not, I think the examples David and Gro-Tsen have come up with conclusively show that there is no geometrically natural construction of the separification.

Inverse limits of arbitrary diagrams don't exist in the category of schemes. How does the argument deal with that? At least it seems to show that $X^s$ exists as a pro-scheme...
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Jonathan WiseJul 29 '13 at 4:19

Yes, this is of course the obvious flaw. But at least we can see that any counterexample must be pathological in some exciting new way, namely that it has infinitely many incompatible maps to separated schemes.
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Will SawinJul 29 '13 at 4:54

Carnahan's suggestion is the natural thing to do in the category of topological spaces, but it's unclear if we may execute it in the category of algebraic spaces, since it's unclear if the projections from the closure of the diagonal down to $X$ are always etale. Note that even for topological spaces, quotienting $X\times X$ out by the closure of the diagonal—i.e. quotienting out $X$ by the relation "$x\sim y$ if there is no pair of open neighbourhoods U of $x$ and $V$ of $y$ such that $U$ and $V$ are disjoint"—doesn't necessarily give a Hausdorff topological space since it's not an equivalence relation: it's not transitive!

But this is a technical problem: the real reason why there shouldn't exist a separification is that separatedness is a global geometric property and it's difficult to replace a scheme with another scheme for which a global geometric property holds, e.g. it's difficult to construct compactifications of schemes, even though that's relatively simple in the category of topological spaces.

At any rate, one way to produce a separification of a scheme is to produce a compactification, e.g. via Nagata's theorem.

Actually, even in the category of topological spaces, if you quotient by the natural relation (two points are equivalent if they aren't "hausdorff-like distinugishable"), you don't always get a hausdorff space. You may have to repeat the procedure many times before you get the hausdorffification of the space.
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Anton GeraschenkoOct 18 '09 at 2:12

2

Being affine is also a global property (every scheme is locally affine), but there does exist an affinification functor.
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Anton GeraschenkoOct 18 '09 at 2:16

Dear Anton, I edited my answer so that it holds under your very correct observations.
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Thanos D. PapaïoannouOct 18 '09 at 21:23

Here is an interesting situation to think about, which might produce a counterexample:

Take two copies of $\mathbb{P}^2$ and consider the standard quadratic (Cremona) birational transformation $\tau\colon (x:y:z) \mapsto (u:v:w)=(yz:xz:xy)$ which restricts to an isomorphism between the open sets $\{xyz\neq 0\}$ of one $\mathbb{P}^2$ and $\{uvw\neq 0\}$ of the other; now call $X$ the scheme obtained by gluing the two $\mathbb{P}^2$ together by identifying these two open sets by means of $\tau$. In other words, $X$ is the projective plane with a triangle doubled, but doubled in two different ways.

If we take the closure of the diagonal $\Delta \subseteq X\times X$, it contains every pair where the first part lies on the line $x=0$ (of the first $\mathbb{P}^2$), say, and the second is the point $v=w=0$ (of the second $\mathbb{P}^2$); this closure is not an equivalence relation, but the equivalence relation it generates identifies all the points of the lines $x=0$, $y=0$, $z=0$ and $u=0$, $v=0$ and $w=0$.

So if we try to "separify" $X$ by somehow quotienting by the closure of the diagonal, we are led to contract the three lines $x=0$, $y=0$ and $z=0$ in $\mathbb{P}^2$ to a single point. Now I believe this cannot be done (except by contracting everything to a point). This suggests that there is no universal morphism from $X$ to a separated scheme (or else that this separification should be a point, which is suspicious).

I think this argument plus rigidity of $\mathbb P^2$ shows that every map from $X$ to a separated scheme is constant, hence $X$ does have a separification - a point, as you suggest.
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Will SawinJul 29 '13 at 5:36